# Solution to Aluminium Block Heating Problem Let's break down the problem and solve it step by step. --- ## **Step 1: Model the Problem and Select the Approach** ### **Given Data** - **Width of slab, L:** .07 m - **Initial temperature, \(T_i\):** 15°C - **Final (target) temperature, \(T_f\):** 500°C - **Hot gas temperature, \(T_{\infty}\):** 550°C - **Convective heat transfer coefficient, \(h\):** 130 W/m²·K - **Thermal conductivity, \(k\):** 231 W/m·K - **Specific heat, \(C_p\):** 1033 J/kg·K - **Density, \(\rho\):** 2702 kg/m³ ### **Approach** Since the aluminium block is being heated by convection and the temperature changes significantly, use the **lumped capacitance method** if the Biot number is less than .1: \[ \text{Bi} = \frac{hL_c}{k} \] where \(L_c = \frac{V}{A}\). For a slab: - \(L_c = \frac{L}{2}\) - \(L = .07 \, \text{m}\) \[ L_c = \frac{.07}{2} = .035\, \text{m} \] \[ \text{Bi} = \frac{130 \times .035}{231} \approx .0197 < .1 \] *The lumped capacitance method is valid.* --- ## **Step 2: Calculate the Time Required** ### **Lumped Capacitance Equation** \[ \frac{T(t) - T_{\infty}}{T_i - T_{\infty}} = \exp\left(-\frac{hA}{\rho V C_p} t\right) \] For a slab of area \(A\) and thickness \(L\): \[ V = A \cdot L \] So, \[ \frac{hA}{\rho V C_p} = \frac{h}{\rho L C_p} \] ### **Insert Values** - \(T(t) = 500^\circ C\) - \(T_\infty = 550^\circ C\) - \(T_i = 15^\circ C\) \[ \frac{500 - 550}{15 - 550} = \frac{-50}{-535} = .0935 \] Take the natural log on both sides: \[ .0935 = \exp\left(-\frac{h}{\rho L C_p} t\right) \] \[ \ln(.0935) = -\frac{h}{\rho L C_p} t \] \[ -2.368 = -\frac{h}{\rho L C_p} t \] \[ 2.368 = \frac{h}{\rho L C_p} t \] Plug in the values: - \(h = 130\) - \(\rho = 2702\) - \(L = .07\) - \(C_p = 1033\) \[ \frac{h}{\rho L C_p} = \frac{130}{2702 \times .07 \times 1033} = \frac{130}{195,595.82} \approx .0006646 \] \[ 2.368 = .0006646 \cdot t \] \[ t = \frac{2.368}{.0006646} \approx 3563\, \text{s} \] --- ## **Final Answer** ### **Time required to heat the aluminium blocks to 500°C:** \[ \boxed{356 \text{ seconds} \approx 59.4 \text{ minutes}} \] --- ## **Graph: Temperature vs. Time** Below is a typical plot for this exponential rise: ```plaintext 550 | * | * Temp| * (°C)| * | * | * | * 15|*_____________________________ 356 t (s) ``` *Curve shows exponential approach from 15°C to 500°C as \(t\) increases.* --- ## **Summary Table** | Step | Calculation | Value | |------------------------------------|-----------------------------|--------------| | Biot Number | \(.0197 < .1\) | Lumped valid | | Time Constant (\(h/\rho LC_p\)) | \(.0006646\) | s\(^{-1}\) | | Time to 500°C (\(t\)) | | 356 s (59.4 min) | --- ### **Example** If you wanted to know how long it takes to reach 400°C, repeat the calculation with \(T(t) = 400°C\): \[ \frac{400 - 550}{15 - 550} = \frac{-150}{-535} = .2804 \] \[ \ln(.2804) = -1.270 \] \[ t = \frac{1.270}{.0006646} = 1,911 \text{ s} \approx 31.8 \text{ min} \] --- **Final Answer:** **It will take approximately 356 seconds (59.4 minutes) to heat the aluminium blocks to 500°C.**